Determine the oxidation number for the indicated element in each of the following compounds: (a) co in licoo2, (b) al in naalh4, (c) c in ch3oh (methanol), (d) n in gan, (e) cl in hclo2, (f) cr in bacro4.

(a) Co = +3

(b) Al = +3

(c) C = -2

(d) N = -3

(e) Cl = -3

(f) Cr = +6

Explanation: Oxidation number: Oxidation number of an element in any particular chemical reaction is the defined as the electrical charge.

(a) The oxidation number of Co in LiCoO₂

Where the oxidation numbers of

Li = +1 , O = -2 and let the oxidation number of Co = x

Therefore,

x + (+1) +  (-2×2) = 0

x + 1 - 4 = 0

x - 3 = 0

x = +3.

Therefore the oxidation number of Co = +3

(b) Al in NaAlH₄

Where Oxidation of Na = +1, H = -1, and let the oxidation number of Al = y

+1 + (y) + (-1×4) = 0

y -4 +1 = 0

y - 3 = 0

y = +3.

Therefore the oxidation number of Al = +3

(c) C in CH₃OH

Where the oxidation number of

H = +1, O = -2, let the oxidation number of C = z

Therefore,

z + (+1×3) + (-2) + (+1) = 0

z + 3 - 2 + 1 = 0

z + 2 = 0

z = -2.

Therefore the oxidation number of C = -2.

(d)  N in GaN

Where Ga = +3 and let the oxidation number of N = a.

Therefore,

a + (+3) = 0

a + 3 = 0

a = -3.

Therefore the oxidation number of N = -3

(e) Cl in HClO₂

Where H = +1, O = -2 and let The oxidation number of Cl = b

+1 + b + (-2×2) = 0

1+b-4 = 0

b + 3 = 0

b = -3

Therefore the oxidation number of Cl = -3

(f) Cr in BaCrO₄

Where Ba = +2, O = -2 and let the oxidation number of Cr = c

+2 + c + (-2×4) = 0

+2+c-8 = 0

c - 6 = 0

c = 6

Therefore the oxidation number of Cr = +6

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